Ok, so I can get through most of this but I can't seem to get the last part... Here is the problem(adsbygoogle = window.adsbygoogle || []).push({});

[tex]xU_x + (y^2+1)U_y = U-1; U(x,x) = e^x[/tex]

Characteristic equations are:

[tex]\frac{dx}{x} = \frac{dy}{y^2+1} = \frac{dU}{U-1}[/tex]

Solving the first and third gives:

[tex]\frac{U-1}{x} = c_1[/tex]

The first and second equation yield:

[tex]tan^{-1}(y) - lnx = c_2[/tex]

Put the two together in the form

[tex]c_1 = f(c_2)[/tex]

[tex]\frac{U-1}{x} = f(tan^{-1}(y) - lnx)[/tex]

Sub in the Cauchy data and you get

[tex]\frac{e^x-1}{x} = f(tan^{-1}(x) - lnx)[/tex]

Now how do I find what my arbitrary function f is? I have spent hours on this. Is there something that relates inverse tan to natural log? Arrggghhhh!

Thanks for any help.

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# First order pde cauchy problem by method of characteristics

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